Assume that a parcel of air is forced to rise up and over a 6000-foot high mountain (shown below). The Initial temperature of the parcel at sea level is 76.5 F, and the lifting condensation level (LCL) of the parcel is 3000 feet. The DAR is 5.5 F/1000' and the SAR is 3.3 F/1000'. Assume the condensation begins at 100% relative humidity and that no evaporation takes place as the parcel descends. Indicate calculated temperatures to one decimal place.
1. Calculate the temp of the parcel at the following elevations as it rises up the windward side of the mountain
A 1000' ________________
B 3000' __________________
C 6000'__________________
2 after the parcel of air has descended dpwn the lee side of the mountain to sea level what is the temperature of the parcel?____________
why?_______________
5.One the winward side of the mountain, should the relative humidity of the parcel change as it rises from 3000 to 6000 feet?__________________
why_____________________
6. As the air rises up the windward side of the mountain
a what is the capacity (saturation mixing rate) of the rising air at 3000 feet?_____________________________g/kg
what is the capacity of the air at 6000 feet?________________________________g/kg
7. what is the capacity of the air after it descended back down to sea level on the lee side of the moutain? ____________
8. Assuming no water vapo is added as the parcel descends down the lee side of the mountain to sea, is the water vapor content (the mixing ration) of the parcel higher or lower than before it began to rise over the mountain?______________________
why_____________________
what is the lifting condensation level of this parcel now? ___________________________feet
1. A 1000': Since the DAR is 5.5 F/1000', the temperature of the parcel at 1000 feet would be 76.5 - (5.5 * 1) = 71.0 F.
B 3000': The DAR is still 5.5 F/1000', so the temperature at 3000 feet would be 76.5 - (5.5 * 3) = 60.0 F.
C 6000': The DAR is still 5.5 F/1000', so the temperature at 6000 feet would be 76.5 - (5.5 * 6) = 38.5 F.
2. After descending down the lee side of the mountain to sea level, the temperature of the parcel would be the same as the initial temperature of 76.5 F. This is because no evaporation takes place during the descent, so there is no change in temperature.
3. The relative humidity of the parcel should change as it rises from 3000 to 6000 feet. As the air rises, it expands and cools, which increases its relative humidity. Therefore, the relative humidity of the parcel would increase.
4. The saturation mixing ratio (capacity) of the rising air at 3000 feet can be calculated using the formula: Mixing ratio at LCL + (SAR * (elevation - LCL)) = Mixing ratio at 3000 feet.
Let's assume the mixing ratio at the LCL is 10 g/kg. Given that the SAR is 3.3 F/1000', we can convert it to F/feet by dividing it by 1000: 3.3 F/1000' = 0.0033 F/foot.
Using the formula, we can calculate the mixing ratio at 3000 feet: 10 + (0.0033 * (3000 - 3000)) = 10 g/kg.
The capacity of the air at 6000 feet would be the same as the mixing ratio at 3000 feet, which is 10 g/kg.
5. The capacity of the air after it descended back down to sea level on the lee side of the mountain would still be 10 g/kg. This is because no evaporation or condensation takes place during the descent, so there is no change in the capacity of the air.
6. Assuming no water vapor is added as the parcel descends down the lee side of the mountain to sea, the water vapor content (mixing ratio) of the parcel would be lower than before it began to rise over the mountain. This is because as the air rises, it cools and condenses, reducing the amount of water vapor in the parcel. As the parcel descends, no evaporation takes place, so the amount of water vapor remains the same.
7. The lifting condensation level (LCL) of this parcel would still be 3000 feet. The LCL represents the level at which the air temperature equals the dew point temperature, so it remains unchanged regardless of the parcel's movements over the mountain. Therefore, the LCL would still be 3000 feet.
To answer these questions, we will follow the steps given in the problem.
1. Calculate the temperature of the parcel at the following elevations as it rises up the windward side of the mountain:
A. 1000' - Start with the initial temperature of 76.5 F and subtract the dry adiabatic rate (DAR) of 5.5 F/1000'. So at 1000', the temperature would be 76.5 - (5.5 * 1) = 71.0 F.
B. 3000' - Start with the temperature at 1000' (which is 71.0 F) and subtract the DAR of 5.5 F/1000' again. So at 3000', the temperature would be 71.0 - (5.5 * 2) = 60.0 F.
C. 6000' - Start with the temperature at 3000' (which is 60.0 F) and subtract the DAR of 5.5 F/1000'. So at 6000', the temperature would be 60.0 - (5.5 * 3) = 43.5 F.
2. After the parcel of air has descended down the lee side of the mountain to sea level, the temperature of the parcel would be same as the initial temperature at sea level, which is 76.5 F. This is because no evaporation takes place as the parcel descends.
3. On the windward side of the mountain, the relative humidity of the parcel would change as it rises from 3000 to 6000 feet. The reason is that as the parcel rises, it expands and cools, and at the lifting condensation level (LCL), it reaches 100% relative humidity and condensation begins, which results in the formation of clouds. This means that the relative humidity is maximum at the LCL and decreases as the parcel continues to rise.
4. Unfortunately, there is no question number 4 mentioned in your statement.
5. The capacity (saturation mixing rate) of the rising air at 3000 feet can be calculated using the formula: Capacity (g/kg) = 20 * (Temperature - Dew Point). Since the parcel is at 100% relative humidity, the temperature and dew point are equal at the LCL. So the capacity at 3000 feet would be Capacity = 20 * (60.0 - 60.0) = 0 g/kg.
The capacity of the air at 6000 feet can be found similarly: Capacity = 20 * (43.5 - 43.5) = 0 g/kg.
7. The capacity of the air after it descended back down to sea level on the lee side of the mountain would still be 0 g/kg since no evaporation takes place during the descent.
8. Assuming no water vapor is added as the parcel descends down the lee side of the mountain, the water vapor content (mixing ratio) of the parcel would be lower than before it began to rise over the mountain. This is because the condensation at the LCL leads to the formation of clouds, which release some of the moisture from the air.
9. The lifting condensation level (LCL) of the parcel is given as 3000 feet.
To solve this question, we need to follow a step-by-step process:
1. Determine the temperature of the parcel at different elevations as it rises up the windward side of the mountain.
A. To find the temperature of the parcel at 1000 feet, we need to know the Dry Adiabatic Rate (DAR) and the Initial Temperature. The DAR is given as 5.5 F/1000', so for every 1000 feet, the temperature will decrease by 5.5 degrees Fahrenheit.
Calculation: Initial Temperature - (Rate x Number of Units)
= 76.5 - (5.5 x 1)
= 71.0 °F
B. Similarly, for 3000 feet, we will use the same calculation:
Calculation: Initial Temperature - (Rate x Number of Units)
= 76.5 - (5.5 x 3)
= 76.5 - 16.5
= 60.0 °F
C. For 6000 feet:
Calculation: Initial Temperature - (Rate x Number of Units)
= 76.5 - (5.5 x 6)
= 76.5 - 33
= 43.5 °F
2. After the parcel of air has descended down the lee side of the mountain to sea level, the temperature of the parcel will be the same as the Initial Temperature since no evaporation or condensation takes place during the descent.
Therefore, the temperature of the parcel at sea level is 76.5°F.
Why? Because it is mentioned in the question that no evaporation takes place as the parcel descends.
3. On the windward side of the mountain, the relative humidity of the parcel should change as it rises from 3000 to 6000 feet.
Why? As the parcel rises, it undergoes adiabatic cooling due to the decrease in pressure. This adiabatic cooling causes the temperature to lower, which increases the relative humidity. So, the relative humidity of the parcel will increase as it rises.
4. The question does not provide information about mixing ratios regarding the air rising up the windward side of the mountain. Therefore, we cannot answer questions 6, 7, and 8 without this information.
5. As for the lifting condensation level (LCL) of the parcel, it is mentioned in the question that the LCL is 3000 feet. So, the lifting condensation level of this parcel remains at 3000 feet.